AGM-consistency and perfect Bayesian equilibrium. Part II: from PBE to sequential equilibrium

Abstract

In (Bonanno, Int J Game Theory 42:567–592, 2013) a general notion of perfect Bayesian equilibrium (PBE) for extensive-form games was introduced and shown to be intermediate between subgame-perfect equilibrium and sequential equilibrium. Besides sequential rationality, the ingredients of the proposed notion are (1) the existence of a plausibility order on the set of histories that rationalizes the given assessment and (2) the notion of Bayesian consistency relative to the plausibility order. We show that a cardinal property of the plausibility order and a strengthening of the notion of Bayesian consistency provide necessary and sufficient conditions for a PBE to be a sequential equilibrium.

Appendix: Proofs

Proof of Lemma 8

We shall construct a full-support common prior that satisfies the properties of Lemma 8. Let $\mathcal {E}$ be the finite collection of equivalence classes of $\precsim$ and let $(E_{1},\ldots ,E_{m})$ be the ordering of $\mathcal {E}$ according to decreasing plausibility, that is, $\forall h,h^{\prime }\in H$, $\forall i,j\in \{1,\ldots ,m\}$, if $h\in E_{i}$ and $h^{\prime }\in E_{j}$ then $h\prec h^{\prime }$ if and only if $i<j$. Let $\mathcal {E}^{+}=\left\{ E\in \mathcal {E}:E\cap D_{\mu }^{+}\ne \varnothing \right\}$ and let $\mathcal {N }=\left\{ \nu _{E}\right\} _{E\in \mathcal {E}^{+}}$ be an arbitrary collection of probability density functions that satisfy the properties of Definition 3. Fix an equivalence class $E_{i}$ and define the function $f_{E_{i}}:H\rightarrow [0,1]$ recursively as follows.

Step 0. For every $h\notin E_{i}$ set $f_{E_{i}}(h)=0$.

Step 1. For every $h\in E_{i}\cap D_{\mu }^{+}$ set $f_{E_{i}}(h)=\nu _{E_{i}}(h),$ where $\nu _{E_{i}}(\cdot )$ is the relevant element of $\mathcal {N}$. Note that, by Property B2 of Definition 3, if $h,ha\in E_{i}\cap D_{\mu }^{+}$ then $f_{E_{i}}(ha)=f_{E_{i}}(h)\times \sigma (a)$.

Step 2. Let $h,ha\in E_{i}$; then (a) if $h\notin D_{\mu }^{+}$ and $ha\in D_{\mu }^{+}$, set $f_{E_{i}}(h)=\frac{f_{E_{i}}(ha) }{\sigma (a)}$ (note that, by P1 of Definition 2, $h\sim ha$ implies $\sigma (a)>0$) and (b) if $h\in D_{\mu }^{+}$ and $ha\notin D_{\mu }^{+}$, set $f_{E_{i}}(ha)=f_{E_{i}}(h)\times \sigma (a)$. Note that, because of Property B2 of Definition 3, the values assigned under Step 2 cannot be inconsistent with the values assigned under Step 1.²⁸

Step 3. After completing Steps 1 and 2, the only histories $h\in E_{i}$ for which $f_{E_{i}}(h)$ has not been defined yet are those that satisfy the following properties: (1) there is no prefix $h^{\prime }$ of h such that $h^{\prime }\in E_{i}\cap D_{\mu }^{+}$ and (2) there is no $h^{\prime }\in E_{i}\cap D_{\mu }^{+}$ such that h is a prefix of $h^{\prime }$. Let $\hat{E}_{i}\subseteq E_{i}$ be the set of such histories (it could be that $\hat{E}_{i}=\varnothing$). A maximal path in $\hat{E}_{i}$ is a sequence $\left\langle h,ha_{1},ha_{1}a_{2},\ldots ,ha_{1}a_{2}\ldots a_{p}\right\rangle$ in $\hat{E}_{i}$ such that $ha_{1}a_{2}\ldots a_{p}$ is a terminal history and there is no $h^{\prime }\in \hat{E}_{i}$ which is a proper prefix of h. Fix an arbitrary maximal path $\left\langle h,ha_{1},\ldots ,ha_{1}a_{2}\ldots a_{p}\right\rangle$ in $\hat{E}_{i}$ and define $f_{E_{i}}(h)=1$ and, for every $i=0,\ldots ,p-1$, $f_{E_{i}}(ha_{1}\ldots a_{i+1})=$ $f_{E_{i}}(ha_{1}\ldots a_{i})\times \sigma (a_{i+1})$ (defining $ha_{0}$ to be h; note that, by P1 of Definition 2 $\sigma (a_{i})>0$ for all $i=1,\ldots ,p$).

By construction, the function $f_{E_{i}}$ satisfies the following property:

$$\begin{aligned} \begin{array}{c} \forall h\in E_{i}, \forall a\in A(h)\text {, if }\sigma (a)>0\text { then (}ha\in E_{i}\text { and)} \\ f_{E_{i}}(ha)=f_{E_{i}}(h)\times \sigma (a)\text { and thus } f_{E_{i}}(ha)\le f_{E_{i}}(h). \end{array} \end{aligned}$$

(9)

Note that Steps 1–3 always assign positive values in (0, 1]; thus if $h\in E_{i}$ then $f_{E_{i}}(h)\in (0,1]$.

Next we show that there exist weights $\lambda _{1},\ldots ,\lambda _{m}\in (0,1)$ such that $\lambda _{1}+\cdots +\lambda _{m}<1$ and, $\forall i,j\in \{1,\ldots ,m\}$, if $h\in E_{i}$, $h^{\prime }\in E_{j}$ and $i<j$ then $\lambda _{j}\times f_{E_{j}}(h^{\prime })\le \lambda _{i}\times f_{E_{i}}(h)$. Fix an arbitrary $\lambda _{1}\in (0,1)$ and let $a=\min _{h\in E_{1}}\left\{ \lambda _{1}\times f_{E_{1}}(h)\right\}$ (clearly, $a\in$ (0, 1)). Let $b=\max _{h\in E_{2}}\left\{ f_{E_{2}}(h)\right\}$ and choose a $\lambda _{2}\in (0,1)$ such that (1) $\lambda _{1}+\lambda _{2}<1$ and (2) $\lambda _{2}\times b\le a$. Then, for every $h\in E_{1}$ and $h^{\prime }\in E_{2}$, $\lambda _{2}\times f_{E_{2}}(h^{\prime })\le \lambda _{1}\times f_{E_{1}}(h)$. Repeat this procedure for choosing a weight $\lambda _{i}$ for every $i\in \{3,\ldots ,m\}.$ Define $\alpha _{i}=\frac{\lambda _{i}}{\lambda _{1}+\cdots +\lambda _{m}}$ and $\bar{\nu }:H\rightarrow [0,1]$ by $\bar{\nu }(h):\sum _{i=1}^{m}\left[ \alpha _{i}\times f_{E_{i}}(h)\right]$. We want to show that, $\forall h\in H$, $\forall a\in A(h)$

$$\begin{aligned} \begin{array}{c} \text {if }ha\in D\text { then (}A\text {) }\bar{\nu }(ha)\le \bar{\nu }(h)\text { and } \\ \text {(}B\text {) if }\sigma (a)>0\text { then }\bar{\nu }(ha)=\bar{\nu } (h)\times \sigma (a). \end{array} \end{aligned}$$

(10)

Fix an arbitrary $ha\in D$. Suppose first that $h\prec ha$. Then, by Property P1 of Definition 2, $\sigma (a)=0,$ so that (B) of (10) is trivially satisfied. Let $E_{i}$ be the equivalence class to which h belongs and $E_{j}$ the equivalence class to which ha belongs, so that $i<j$. Then $\bar{\nu }(h)=\alpha _{i}\times f_{E_{i}}(h)$ and $\bar{\nu }(ha)=\alpha _{j}\times f_{E_{j}}(ha)$ and thus, since $\lambda _{j}\times f_{E_{j}}(ha)\le \lambda _{i}\times f_{E_{i}}(h)$ , dividing both sides by $(\lambda _{1}+\cdots +\lambda _{m})$ we get that $\bar{ \nu }(ha)\le \bar{\nu }(h)$. Suppose now that $h\sim ha$ (so that, by Property P1 of Definition 2, $\sigma (a)>0$). Let $E_{i}$ be the equivalence class to which both h and ha belong; then, $\bar{\nu } (h)=\alpha _{i}\times f_{E_{i}}(h)$ and $\bar{v}(ha)=\alpha _{i}\times f_{E_{i}}(ha)$ and thus (10) follows from (9).

Finally, define $\nu :D\rightarrow (0,1]$ by $\nu (h)=\frac{\bar{\nu }(h)}{ \sum _{h^{\prime }\in D}\bar{\nu }(h^{\prime })}$. Then it follows from (10) that if $ha\in D$ then (A) $\nu (ha)\le \nu (h)$ and (B) if $\sigma (a)>0$ then $\nu (ha)=\nu (h)\times \sigma (a)$. It only remains to prove that $\nu$ is a common prior of $\mathcal {N}=\left\{ \nu _{E}\right\} _{E\in \mathcal {E}^{+}}$, that is, that, $\forall i\in \{1,\ldots ,m\}$ and $\forall h\in E_{i}\cap D_{\mu }^{+}$, $v(h~|~E_{i}\cap D_{\mu }^{+})\overset{ def}{=}\frac{\nu (h)}{\sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}~\nu (h^{\prime })}=\nu _{E_{i}}(h)$, where $\nu _{E_{i}}(\cdot )$ is the relevant element of $\mathcal {N}$. Now, $\frac{\nu (h)}{\sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}~\nu (h^{\prime })}=\frac{\nu (h)~{\large \times } ~\sum _{h^{\prime \prime }\in D}~\bar{\nu }(h^{\prime \prime })}{ \sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}~\nu (h^{\prime })~{\large \times }~\sum _{h^{\prime \prime }\in D}~\bar{\nu }(h^{\prime \prime })}=\frac{ \bar{\nu }(h)}{\sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}\left[ \nu (h^{\prime })~{\large \times }~\sum _{h^{\prime \prime }\in D}~\bar{\nu } (h^{\prime \prime })\right] }=\frac{\bar{\nu }(h)}{\sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}~\bar{\nu }(h^{\prime })}=\frac{\alpha _{i}~\times ~f_{E_{i}}(h)}{\sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}~\left[ \alpha _{i}~\times ~f_{E_{i}}(h^{\prime })\right] }=\frac{\alpha _{i}~\times ~\nu _{E_{i}}(h)}{\sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}~\left[ \alpha _{i}~\times ~\nu _{E_{i}}(h^{\prime })\right] }=\frac{\alpha _{i}~\times ~\nu _{E_{i}}(h)}{\alpha _{i}~\times ~\sum _{h^{\prime }\in E_{i}\cap D_{\mu }^{+}}\nu _{E_{i}}(h^{\prime })}=\nu _{E_{i}(h)}$ since, by construction, for all $h\in E_{i}\cap D_{\mu }^{+}$, $f_{E_{i}}(h)=\nu _{E_{i}}(h)$ and $Supp(\nu _{E_{i}})=E_{i}\cap D_{\mu }^{+}$. $\square$

In order to prove Proposition 11 we will exploit the characterization of sequential equilibrium given in Perea et al. (1997). First some notation and terminology. Let A be the set of actions (as in Bonanno (2013), we assume that no action is available at more than one information set, that is, if $h^{\prime }\notin I(h)$ then $A(h^{\prime })\ne A(h)$). If h is a history, we denote by $A_{h}$ the set of actions that occur in history h (thus while h is a sequence of actions, $A_{h}$ is the set of actions in that sequence; note that, for every history h, $A_{h}\ne \varnothing$ if and only if $h\ne \emptyset$). Given an assessment $(\sigma ,\mu )$ we denote by $A^{0}=\{a\in A:\sigma (a)=0\}$ the set of actions that are assigned zero probability by the strategy profile $\sigma$ . Recall that $D_{\mu }^{+}$ denotes the set of decision histories to which $\mu$ assigns positive probability ($D_{\mu }^{+}=\left\{ h\in D:\mu (h)>0\right\}$) and that $h^{\prime }\in I(h)$ mans that h and $h^{\prime }$ belong to the same information set. A pseudo behavior strategy profile (PBSP) is a generalization of the notion of behavior strategy profile that allows the sum of the “probabilities” over the actions at an information set to be larger than 1, that is, a PBSP is a function $\bar{\sigma }:A\rightarrow [0,1].$ A PBSP $\bar{\sigma }$ is a completely mixed extension of a behavior strategy profile $\sigma$ if, $\forall a\in A$, (1) $\bar{ \sigma }(a)>0$ and (2) if $\sigma (a)>0$ then $\bar{\sigma }(a)=\sigma (a)$. Given a PBSP $\bar{\sigma }$, for every history h let $\mathbb {P}_{\bar{ \sigma }}(h)=\left\{ \begin{array}{ll} 1 &{} \quad \text {if }h=\emptyset \\ \bar{\sigma }(a_{1})\times \cdots \times ~\bar{\sigma }(a_{m}) &{} \quad \text {if } h=a_{1}\ldots a_{m} \end{array} \right. .$ The following proposition is proved in Perea et al. (1997) (see also (Perea 2001, p. 74 and Streufert 2007).

Proposition 13

(Perea et al. 1997, Theorem 3.1, p. 241) Fix an extensive game and let $(\sigma ,\mu )$ be an assessment. Then (A) and (B) below are equivalent:

(A)

• (A.1) There exists a function $\varepsilon :A^{0}\rightarrow (0,1)$ such that, $\forall h,h^{\prime }\in D$ with $h^{\prime }\in I(h)$,

  (A.1a) If $h,h^{\prime }\in D_{\mu }^{+}$ then $\prod \nolimits _{a\in A^{0}\cap A_{h}}~\varepsilon (a)=\prod \nolimits _{a\in A^{0}\cap A_{h^{\prime }}}~\varepsilon (a)$, and

  (A.1b) if $h\in D_{\mu }^{+}$ and $h^{\prime }\notin D_{\mu }^{+}$ then $\prod \nolimits _{a\in A^{0}\cap A_{h}}~\varepsilon (a)>\prod \nolimits _{a\in A^{0}\cap A_{h^{\prime }}}~\varepsilon (a)$, and

• (A.2) there is a PBSP $\bar{\sigma }$ which is a completely mixed extension of $\sigma$ and is such that,$\forall h,h^{\prime }\in D_{\mu }^{+}$ with $h^{\prime }\in I(h)$, $\frac{\mathbb {P}_{\bar{\sigma }}(h)}{\mathbb {P}_{\bar{\sigma } }(h^{\prime })}=\frac{\mu (h)}{\mu (h^{\prime })}$.

• (B) $(\sigma ,\mu )$ is KW-consistent.

Proof of Proposition 11

$(I)\Rightarrow (II)$. Let $(\sigma ,\mu )$ be a perfect Bayesian equilibrium which is rationalized by a choice measurable plausibility order $\precsim$ and is uniformly Bayesian relative to it. We need to show that $(\sigma ,\mu )$ is a sequential equilibrium. Since $(\sigma ,\mu )$ is a perfect Bayesian equilibrium, it is sequentially rational and thus we only need to show that $(\sigma ,\mu )$ is KW-consistent. We shall use choice measurability (Definition 7) to obtain the function $\varepsilon$ of Proposition 13 (a similar argument can be found in Streufert (2012)) and the full-support common prior $\nu$ of Definition 9 to obtain the PBSP $\bar{\sigma }$ .

By hypothesis $\precsim$ is choice measurable. Fix a cardinal integer-valued representation F of $\precsim$ and normalize it so that $F(\emptyset )=0$.²⁹ For every action $a\in A$, define $\varepsilon (a)=e^{\left[ F(h)-F(ha)\right] }$ for some h such that $a\in A(h)$. By definition of choice measurability, if $h^{\prime }\in I(h)$ then $F(h)-F(ha)=F(h^{\prime })-F(h^{\prime }a)$ and thus the function $\varepsilon (a)$ is well defined. Furthermore, if $a\in A^{0}$ (that is, $\sigma (a)=0$) it follows from P1 of Definition 2 that $h\prec ha$ and thus $F(h)-F(ha)<0$ so that $0<\varepsilon (a)<1$, while if $a\notin A^{0}$ (that is, $\sigma (a)>0$) then, by P1 of Definition 2, $h\sim ha$ and thus $F(h)-F(ha)=0$ so that $\varepsilon (a)=1$; hence,

$$\begin{aligned} \text {if }A^{0}\cap A_{h}\ne \varnothing ,\text { then }\prod \nolimits _{a\in A_{h}}~\varepsilon (a)=\prod \nolimits _{a\in A^{0}\cap A_{h}}~\varepsilon (a). \end{aligned}$$

(11)

We want to show that the restriction of $\varepsilon (\cdot )$ to $A^{0}$ satisfies (A.1) of Proposition 13. Fix an arbitrary history $h=a_{1}a_{2}\ldots a_{m}$. Since $\left[ F(\emptyset )-F(a_{1})\right] +\left[ F(a_{1})-F(a_{1}a_{2})\right] +\cdots +\left[ F(a_{1}\ldots a_{m-1})-F(h)\right] =-F(h)$ (recall that $F(\emptyset )=0$), it follows that,

$$\begin{aligned} \forall h\in H\backslash \{\emptyset \}, \prod \nolimits _{a\in A_{h}}~\varepsilon (a)=e^{-F(h)}. \end{aligned}$$

(12)

Fix arbitrary $h,h^{\prime }\in D\backslash \{\emptyset \}$ with $h^{\prime }\in I(h)$ and $h\in D_{\mu }^{+}$ (that is, $\mu (h)>0$).³⁰ Suppose first that $h^{\prime }\in D_{\mu }^{+}$. Then, by P2 of Definition 2, $h\sim h^{\prime }$ and thus $F(h)=F(h^{\prime })$, so that, by (12), $\prod \nolimits _{a \in A_{h}}~\varepsilon (a)=\prod \nolimits _{a\in A_{h^{\prime }}}~\varepsilon (a).$ Thus, by (11), (A.1a) of Proposition 13 is satisfied. Suppose now that $h^{\prime }\notin D_{\mu }^{+}$. Then, by P2 of Definition 2, $h\prec h^{\prime }$ and thus $F(h)<F(h^{\prime })$ so that $e^{-F(h^{\prime })}<e^{-F(h)}$ and thus, by (12), $\prod \nolimits _{a\in A_{h}}~\varepsilon (a)>\prod \nolimits _{a\in A_{h^{\prime }}}~\varepsilon (a)$. Thus, by (11), (A.1b) of Proposition 13 is also satisfied.

Next we prove (A.2). Denote by $\bar{A}$ the set of “non-terminal actions” , that is, $\bar{A}=\{a\in A:ha\in D$ for some h with $a\in A(h)\}$. Let $\nu$ be a uniform full-support common prior (Definition 9). Define $\bar{\sigma }:\bar{A}\rightarrow (0,1]$ as follows: $\bar{\sigma }(a)=\frac{\nu (ha)}{\nu (h)}$ for some h such that $a\in A(h)$ and $ha\in D$. By Property UB2 of Definition 9, if $h^{\prime }\in I(h)$ then $\frac{\nu (h^{\prime }a)}{\nu (h^{\prime })}=\frac{\nu (ha)}{\nu (h)}$ and thus $\bar{\sigma }$ is well defined; furthermore, since $\nu (h)>0$ for all $h\in D$, $\bar{\sigma } (a)>0.$ By (A) of Property UB1 of Definition 9, $\bar{\sigma }(a)\le 1.$ Finally, by (B) of Property UB1 of Definition 9, if $\sigma (a)>0$ then $\bar{\sigma }(a)=\sigma (a)$. Thus $\bar{\sigma }$ is a PBSP which is a completely mixed extension of $\sigma$. We need to show that $\forall h,h^{\prime }\in D_{\mu }^{+}$ with $h^{\prime }\in I(h)$, $\frac{\mathbb {P}_{\bar{\sigma }}(h)}{\mathbb {P}_{\bar{\sigma }}(h^{\prime }) }=\frac{\mu (h)}{\mu (h^{\prime })}$. If $h=\emptyset$ it is trivially true because $h^{\prime }=h$ and $\mathbb {P}_{\bar{\sigma }}(\emptyset )=\mu (\emptyset )=1$. Fix arbitrary $h,h^{\prime }\in D_{\mu }^{+}\backslash \{\emptyset \}$ with $h^{\prime }\in I(h).$ Let $h=a_{1}a_{2}\ldots a_{p}$ ($p\ge 1$) and $h^{\prime }=b_{1}b_{2}\ldots b_{r}$ ($r\ge 1$). By definition of $\bar{\sigma }$, $\mathbb {P}_{\bar{\sigma }}(h)=\bar{\sigma }(a_{1})\times \bar{ \sigma }(a_{2})\times \cdots \times ~\bar{\sigma }(a_{p})=\frac{\nu (a_{1})}{\nu (\emptyset )}\times \frac{\nu (a_{1}a_{2})}{\nu (a_{1})}\times \cdots \times \frac{\nu (h)}{\nu (a_{1}a_{2}\ldots a_{p-1})}=\frac{\nu (h)}{\nu (\emptyset )}$ . Similarly, $\mathbb {P}_{\bar{\sigma }}(h^{\prime })=\frac{\nu (h^{\prime }) }{\nu (\emptyset )}$. Thus $\frac{\mathbb {P}_{\bar{\sigma }}(h)}{\mathbb {P}_{ \bar{\sigma }}(h^{\prime })}=\frac{\nu (h)}{\nu (h^{\prime })}$. Dividing numerator and denominator of the right-hand-side by $\mathop {\sum }\nolimits _{h^{\prime \prime }\in E_{i}\cap D_{\mu }^{+}}\nu (h^{\prime \prime })$ and using the fact that (since $\nu$ is a common prior) $\frac{\nu (h)}{ \mathop {\sum }\nolimits _{h^{\prime \prime }\in E_{i}\cap D_{\mu }^{+}}\nu (h^{\prime \prime })}=\nu _{E_{i}}(h)$ and $\frac{\nu (h^{\prime })}{ \mathop {\sum }\nolimits _{h^{\prime \prime }\in E_{i}\cap D_{\mu }^{+}}\nu (h^{\prime \prime })}=\nu _{E_{i}}(h^{\prime })$, where $E_{i}$ is the equivalence class to which both h and $h^{\prime }$ belong, we get that $\frac{\mathbb { P}_{\bar{\sigma }}(h)}{\mathbb {P}_{\bar{\sigma }}(h^{\prime })}=\frac{\nu _{E_{i}}(h)}{\nu _{E_{i}}(h^{\prime })}$; now, dividing numerator and denominator of the right-hand-side by $\mathop {\sum }\nolimits _{h^{\prime \prime }\in I(h)}\nu _{E_{i}}(h^{\prime \prime })$ and using the fact that, by B3 of Definition 3, $\frac{\nu _{E_{i}}(h)}{\mathop {\sum }\nolimits _{h^{\prime \prime }\in I(h)}\nu _{E_{i}}(h^{\prime \prime })}=\mu (h)$ and $\frac{\nu _{E_{i}}(h^{\prime })}{\mathop {\sum }\nolimits _{h^{\prime \prime }\in I(h)}\nu _{E_{i}}(h^{\prime \prime })}=\mu (h^{\prime })$, we obtain $\frac{\mathbb {P} _{\bar{\sigma }}(h)}{\mathbb {P}_{\bar{\sigma }}(h^{\prime })}=\frac{\mu (h)}{ \mu (h^{\prime })}$, so that (A.2) of Proposition 13 also holds. Hence, by Proposition 13, $(\sigma ,\mu )$ is KW-consistent.

$(II)\Rightarrow (I)$. Let $(\sigma ,\mu )$ be a sequential equilibrium. That $(\sigma ,\mu )$ is rationalized by a choice measurable plausibility order $\precsim$ and is Bayesian relative to it was proved in Bonanno (2013).³¹ Thus we only need to show that it is uniformly Bayesian. By Proposition 13, there exists a completely mixed PBSP $\bar{\sigma }$ that extends $\sigma$ and is such that,

$$\begin{aligned} \forall h,h^{\prime }\in D_{\mu }^{+}\quad \text { with }h^{\prime }\in I(h),~~ \frac{\mathbb {P}_{\bar{\sigma }}(h)}{\mathbb {P}_{\bar{\sigma }}(h^{\prime })}= \frac{\mu (h)}{\mu (h^{\prime })}. \end{aligned}$$

(13)

Define $\bar{\nu }:D\rightarrow (0,1]$ recursively as follows: $\nu (\emptyset )=1$ and, if $a\in A(h)$ and $ha\in D$, $\bar{\nu }(ha)=\bar{\nu } (h)\times \bar{\sigma }(a)$. Since, $\forall a\in A$, $\bar{\sigma }(a)\in (0,1]$ and $\bar{\sigma }(a)=\sigma (a)$ whenever $\sigma (a)>0$, it follows that

$$\begin{aligned} \begin{array}{c} \text {if }a\in A(h)\text { and }ha\in D\text {, then (}A\text {) }\bar{\nu } (ha)\le \bar{\nu }(h)\text { and} \\ \text {(}B\text {) if }\sigma (a)>0\text { then }\bar{\nu }(ha)=\bar{\nu } (h)\times \sigma (a). \end{array}\nonumber \\ \end{aligned}$$

(14)

Define the probability density function $\nu :D\rightarrow (0,1]$ by $\nu (h)=\frac{\bar{\nu }(h)}{\mathop {\sum }\nolimits _{h^{\prime }\in D}\bar{\nu }(h^{\prime })}.$ Then, by (14), $\nu$ satisfies Property UB1 of Definition 9. Furthermore, if $a\in A(h)$, h and $h^{\prime }$ belong to the same information set and $ha,h^{\prime }a\in D$, then $\frac{\bar{\nu }(ha)}{\bar{\nu }(h^{\prime }a)}=\frac{\bar{\nu } (h)\times \bar{\sigma }(a)}{\bar{\nu }(h^{\prime })\times \bar{\sigma }(a)}= \frac{\bar{\nu }(h)}{\bar{\nu }(h^{\prime })}$ and thus, dividing numerator and denominator by $\mathop {\sum }\nolimits _{h^{\prime \prime }\in D}\bar{\nu } (h^{\prime \prime })$, we get that $\nu$ satisfies Property UB2 of Definition 9.

Furthermore, as shown above,

$$\begin{aligned} \forall h\in D\backslash \{\emptyset \},~\mathbb {P}_{\bar{\sigma }}(h)\overset{def}{=}\mathop {\prod }\limits _{a\in A_{h}}\bar{\sigma }(a)=\frac{\bar{\nu }(h)}{\bar{ \nu }(\emptyset )}=\bar{\nu }(h)\text { (since }\nu (\emptyset )=1\text {),} \end{aligned}$$

(15)

it follows from (13) and (15) that

$$\begin{aligned} \forall h,h^{\prime }\in D_{\mu }^{+}\text { with }h^{\prime }\in I(h),~~ \frac{{\bar{\nu }}(h)}{{\bar{\nu }}(h^{\prime })}=\frac{\mu (h)}{ \mu (h^{\prime })}\quad \text { and thus }\frac{{\nu }(h)}{{\nu } (h^{\prime })}=\frac{\mu (h)}{\mu (h^{\prime })}. \end{aligned}$$

(16)

Fix an arbitrary equivalence class E of the plausibility order that rationalizes $(\sigma ,\mu )$ such that $E\cap D_{\mu }^{+}\ne \varnothing$ and define $\nu _{E}:H\rightarrow [0,1]$ as follows:

$$\begin{aligned} \nu _{E}(h)=\left\{ \begin{array}{ll} \frac{\nu (h)}{\sum _{h^{\prime }\in E\cap D_{\mu }^{+}}\nu (h^{\prime })} &{} \quad \hbox {if}\,h\in E\cap D_{\mu }^{+}\\ 0 &{} \quad \hbox {if}\,h\notin E\cap D_{\mu }^{+}. \end{array} \right. \end{aligned}$$

(17)

By construction $\nu _{E}$ satisfies Property B1 of Definition 3 and, by UB1 of Definition 9 (proved above), $\nu _{E}$ satisfies also B2 of Definition 3 (recall that if $h,h^{\prime }\in E$ with $h^{\prime }=ha_{1}\ldots a_{m}$ then, by P1 of Definition 2, $\sigma (a_{i})>0$ for all $i=1,\ldots ,m$). It only remains to prove that Property B3 of Definition 3 is satisfied, namely that if $h\in E\cap D_{\mu }^{+}$ then, for every $h^{\prime }\in I(h)$, $\mu (h^{\prime })=\frac{\nu _{E}(h^{\prime })}{ \sum _{h^{\prime \prime }\in I(h)}\nu _{E}(h^{\prime \prime })}$. Number the elements of $E\cap D_{\mu }^{+}$ from 1 to m in such a way that $h_{1}=h$ and the first p elements belong to $I(h_{1})$ and the remaining elements (if any) do not belong to $I(h_{1})$, that is, $E\cap D_{\mu }^{+}=\left\{ h_{1},\ldots ,h_{p},h_{p+1},\ldots ,h_{m}\right\}$ with $h_{1}=h$, $I(h_{1})\cap E\cap D_{\mu }^{+}=\{h_{1},\ldots ,h_{p}\}$ and, for $i>p,$ $h_{i}\notin I(h_{1})$. We shall prove that

$$\begin{aligned} \frac{\nu _{E}(h_{1})}{\sum _{h^{\prime \prime }\in I(h_{1})}\nu _{E}(h^{\prime \prime })}=\mu (h_{1}). \end{aligned}$$

(18)

The proof for $1<j\le p$ is similar. By (16), for every $j=1,\ldots ,m$, $\frac{{\nu }(h_{j})}{{\nu }(h_{1})}= \frac{\mu (h_{j})}{\mu (h_{1})}$. Thus

$$\begin{aligned} \frac{\sum _{j=1}^{p}{\nu }(h_{j})}{{\nu }(h_{1})}=\frac{ \sum _{j=1}^{p}{\mu }(h_{j})}{{\mu }(h_{1})}. \end{aligned}$$

(19)

By definition of $\mu$, $\sum _{j=1}^{p}{\mu } (h_{j})=1$ (since, for any $h^{\prime }\in I(h_{1})$ that does not belong to $E\cap D_{\mu }^{+}$, $\mu (h^{\prime })=0$: recall that, by Property P2 of Definition 2, if $h^{\prime }\in I(h_{1})$ is such that $\mu (h^{\prime })>0$ then $h^{\prime }\sim h$, that is, $h^{\prime }\in E$). Hence $\frac{{\nu }(h_{1})}{\sum _{j=1}^{p}{\nu }(h_{j})}=\mu (h_{1}).$ By (17), dividing numerator and denominator of left-hand-side by $\sum _{i=1}^{m}{\nu }(h_{i})$ we obtain

$$\begin{aligned} \frac{{\nu }_{E}(h_{1})}{\sum _{j=1}^{p}{\nu }_{E}(h_{j})}=\mu (h_{1}) \end{aligned}$$

(20)

Since, by (17), for any $h^{\prime }\in I(h_{1})$ that does not belong to $E\cap D_{\mu }^{+}$, $\nu _{E}(h^{\prime })=0$, $\sum _{h^{\prime \prime }\in I(h_{1})}\nu _{E}(h^{\prime \prime })=\sum _{j=1}^{p}{\nu }_{E}(h_{j})$. Thus (20) yields the desired (18). Since, by construction, $\nu$ is a full-support common prior to the collection of probability density functions $\nu _{E}$ given in (17), which have been shown to satisfy the properties of Definition 3, the proof that $(\sigma ,\mu )$ is uniformly Bayesian is complete. $\square$